Pólya and Szegő, Part I, Ch. 4, 174.

Solution 1:

I'll try a short answer, then I can make it longer and go into more details. Let $x_n$ be a sequence such that $x_{n+1}=f(x_n)$ with $f(x) = x-ax^{k}+o(x^k)$.

I'll provide you another proof, or say, a manner of thinking it :

We have $x_{n+1}-x_n = f(x_n)-x_n \sim a x_n^{k}$. The "equivalent" for functions is the ODE $y'=a y^{k}$ which reads $(y^{1-k})' = a(1-k)$.

Then the "discrete equivalent" of that is that we can expect that : $$x_{n+1} ^{1-k} - x_n^{1-k} \to a(1-k).$$ And then we prove this by just the same way you did.

Please not that I do not pretend of a truly new proof, the most interesting part of it is the method. For example it apply to the problem of finding an equivalent (or asymptotic development) of $x_{n+1}=x_{n}^2 \log x_n$ which is more tricky ;)